An electric bulb is designed to consume `55 W` when operated at 110 volts. It is connected to a `220 V, 50 Hz` line through a choke coil in series. What should be the inductance of the coil for which the bulb gets correct voltage?

Answer

To solve the problem, we need to determine the inductance of the choke coil that allows the bulb to operate correctly at 110 volts when connected to a 220 V supply. Here’s a step-by-step solution:

### Step 1: Calculate the resistance of the bulb
The power consumed by the bulb is given as 55 W when operated at 110 V. We can use the formula for resistance (R) based on power (P) and voltage (V):

\[ R = \frac{V^2}{P} \]

Substituting the values:

\[ R = \frac{110^2}{55} = \frac{12100}{55} = 220 \, \Omega \]

### Step 2: Calculate the current through the bulb
Using the power formula, we can find the current (I) when the bulb operates at 110 V:

\[ I = \frac{P}{V} = \frac{55}{110} = 0.5 \, A \]

### Step 3: Relate the current to the total voltage in the AC circuit
In an AC circuit, the total voltage (V) is related to the impedance (Z) and the current (I):

\[ V = I \cdot Z \]

Given that the supply voltage is 220 V, we have:

\[ 220 = 0.5 \cdot Z \implies Z = \frac{220}{0.5} = 440 \, \Omega \]

### Step 4: Use the impedance formula
The impedance (Z) in an R-L circuit is given by:

\[ Z = \sqrt{R^2 + X_L^2} \]

Where \(X_L\) is the inductive reactance given by:

\[ X_L = \omega L = 2 \pi f L \]

With \(f = 50 \, Hz\), we can express \(X_L\) as:

\[ X_L = 2 \pi (50) L = 100 \pi L \]

Now substituting into the impedance formula:

\[ 440 = \sqrt{220^2 + (100 \pi L)^2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:

\[ 440^2 = 220^2 + (100 \pi L)^2 \]

Calculating \(440^2\) and \(220^2\):

\[ 193600 = 48400 + (100 \pi L)^2 \]

### Step 6: Solve for \(L\)
Rearranging the equation:

\[ (100 \pi L)^2 = 193600 - 48400 = 145200 \]

Taking the square root:

\[ 100 \pi L = \sqrt{145200} \]

Calculating \(\sqrt{145200}\):

\[ 100 \pi L = 381.5 \implies L = \frac{381.5}{100 \pi} = \frac{381.5}{314.16} \approx 1.214 \, H \]

### Final Result
Thus, the inductance \(L\) required for the choke coil is approximately:

\[ L \approx 1.2 \, H \]

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